# Algebraic de Rham Cohomology of Projective Space over $\mathbb{C}$

I am looking to understand algebraic de Rham cohomology a bit better, and I realized that I don't really understand something that should be quite elementary. Take for example $$\mathbb{P}^1_{\mathbb{C}}$$. We know that the algebraic de Rham Cohomology $$\mathbb{H}^2(\mathbb{P}^1_{\mathbb{C}},\Omega^{\bullet}_{alg})$$ should be one dimensional (for example since it is isomorphic to singular cohomology for this space). But could one possibly write down a representative for a cohomology class that generates it?

The definition of algebraic de Rham cohomology is quite abstract, with taking a resolution of the complex $$\Omega^{\bullet}_{alg}$$ etc. but I was wondering if in simple, but still non-affine, cases something explicit could be said about it? Perhaps another example could be $$Q \subset \mathbb{P}^3_{\mathbb{C}}$$ - a quadric of dimesion 2 in projective 3-space, which has non-trivial second cohomology as well. Any leads would be great.

• It's not so hard to take an acyclic resolution of $\Omega^\cdot_\text{alg}$. Resolve $\mathcal{O}$ and $\mathcal{\Omega}$ separately (use the Cech resolution attached to the standard cover), make horizontal arrows, and then take the total complex. See the calculations on pp. 7-8 in Kedlaya's expository article here: swc.math.arizona.edu/aws/2007/KedlayaNotes11Mar.pdf Mar 22, 2019 at 5:32
• The answer you get for a projective algebraic curve, say one that you've covered by two affine opens $U$ and $V$, is that $\mathbb{H}^2_\text{dR}$ is holomorphic one-forms on $U \cap V$, mod the subgroup generated by exact forms and forms that are restrictions from holomorphic forms on all of $U$ or all of $V$. The map taking such a form to the sum of its residue on $U^c$ is an identification of this cohomology group with $k$. Mar 22, 2019 at 5:39
• @hunter after putting my answer below, (and going to bed) I did think, gee really, obviously the correct and desired answer was something as you suggest. If you wish to add a version 3 to my answer - please do. As I wrote below, I don't mind making it (once it ends up being 1/2 way decent!) a community answer. Mar 22, 2019 at 13:09
• @peterag I think your answer is also correct and desired in that it uses the analytic machinery available over $\mathbb{C}$, which OP explicitly asked about. I don't think it should be made community wiki -- it's good enough already! With that in mind, I am writing up the algebraic answer in a separate answer. Mar 22, 2019 at 13:12

## 2 Answers

For a sheaf $$\mathcal{F}$$ on $$\mathbb{P}^1$$ and an open $$U$$ in $$\mathbb{P}^1$$, write $$\mathcal{F}_U$$ for the sheaf that takes an open $$V$$ to $$\mathcal{F}(U \cap V)$$. (This is $$\iota_* \mathcal{F}|_U$$ in fancy language.)

Let $$U = \mathbb{P}^1 \setminus \{\infty\}$$ and $$V= \mathbb{P}^1 \setminus \{0\}$$. Then we get an acyclic resolution of the complex

$$0 \to \mathcal{O} \to \ \Omega \to 0$$

on $$\mathbb{P}^1$$ by taking the complex $$0 \to \mathcal{O}_U \oplus \mathcal{O}_V \to \mathcal{O}_{U \cap V} \oplus \Omega_U \oplus \Omega_V \to \Omega_{U \cap V} \to 0.$$ All the sheaves involved are acyclic because they're supported on affines. The maps are, respectively: $$(f, g) \mapsto (f - g, df, dg)$$ and $$(\phi, \omega, \eta) \mapsto d\phi + \omega - \eta$$.

(Three notes: one, a little abuse of notation in defining these maps, we've suppressed restrictions of a function or form from their domain to a smaller domain from the notation.

Two, every smooth projective curve can be covered by two affines, and the same resolution works for every smooth projective curve.

Three, it is reasonable to ask how you would ever think of this resolution. The point is that we've taken the Cech resolution of $$\Omega$$ and $$\mathcal{O}$$ separately, then taken the total complex of the associated double complex. See Kedlaya's article that I left in a comment for more detials, and a proof this always works.)

Let's say we're working in characteristic zero (the definitions make sense in characteristic $$p$$ but are not typically used since the cohomology theory doesn't satisfy Weil's axioms). Taking global sections and computing homology, we get $$H^0(\mathbb{P}^1) = \{(f, g) \in \mathcal{O}(U) \oplus \mathcal{O}(V) \ | \ (f - g, df, dg) = (0, 0, 0)\}.$$ Since in characteristic zero $$df = 0$$ implies $$f$$ is constant, this shows $$H^0$$ is one-dimensional generated by $$(\lambda, \lambda)$$. (Same proof for any smooth projective curve.)

Now the interesting case, $$H^1(\mathbb{P}^1) = \frac{ \{ (\phi, \omega, \eta) \in \mathcal{O}(U \cap V)\oplus \Omega(U)\oplus \Omega{O}(V)\ | \ d\phi = \omega - \eta\} } { \big\{(\phi, \omega, \eta): \phi \text{ can be written as a difference } f - g \text{ where} f \text{ extends to a function on } U \\ \ \ \ \ \text{ and } g \text{ extends to a function on } V \text{ and } df = \omega \text{ and } dg = \eta \big\}}.$$

This seems verbose but in fact is $$0$$, which you can see by letting $$z$$ be a uniformizer at $$0$$ and writing everything in terms of Laurent polynomials in $$z$$, then explicitly taking antiderivatives. The key is that you can't get any $$dz/z$$ terms as these are neither regular at $$0$$ or $$\infty$$, but you can antidifferentiate anything else.

Finally, $$H^2(\mathbb{P}^1)$$ identifies with forms supported on $$\Omega_{U \cap V}$$ mod exact forms, and mod the image of the restriction maps from $$\Omega_U$$ or $$\Omega_V$$. Again using Laurent polynomials, we see this space is generated by the image of $$\frac{dz}{z}$$.

In what follows, always assume $$X$$ to be a Riemann surface - so that one should expect $${\mathbb H}^2 (X,\Omega^\bullet)$$ to be Hodge theory's $$H^{(1,1)}$$.

Version 1: In the complex topology, $$(\Omega^\bullet,\partial)$$ is quasi-isomorphic to the de Rahm complex of sheaves $$({\cal A}^\bullet, d)$$ (with $$d = \partial +\overline\partial$$): on the one hand, we have the $$d$$-Poincare lemma; on the other, locally we can integrate any holomorphic differential form to obtain a holomorphic function - which is good enough for Riemann surfaces. [According to Griffith and Harris, p 448, the corresponding statement ($$\partial$$-Poincare) is true in higher dimensions/degrees.]

Be that as it may, we may therefore calculate the algebraic de Rham using de Rham's de Rham... Write $${\cal A}^k= \sum_ {p+q =k} {\cal A}^{(p,q)}$$, where $${\cal A}^{(p,q)}$$ is the (soft) sheaf of $$(p,q)$$ differential forms, and write $$d = \partial + \overline\partial$$. The spectral sequence (filtration on the $$p$$) degenerates at the $$E_1$$ term, with $$E_1^{p,q} = H^q_{\overline\partial}(A^{(p,\bullet)})$$, where $$A^{(p,q)}$$ are the global sections of $${\cal A}^{(p,q)}$$. Hence, in the case of $$X$$ a Riemann surface, one has the identification $${\mathbb H}^2(X, \Omega^{\bullet}_{alg}) = H^1_{\overline\partial}(A^{(1,\bullet)} )= H^1(X, \Omega^1)$$. In this analytic/calculus setting, one can choose as generator for $$H^{(1,1)}$$ the Chern class of the line bundle $${\cal O}(\infty)$$, or the Fubini-Study Kaehler form...

Version 2 ( more algebraic, and/but weaker): If $$C$$ is a complex, write $$C[k]$$ for the same complex, but shifted so that $$C[k]^n = C^{k+n}$$. Then, in the case of a curve $$X$$, on has an exact sequence of (sheaf) complexes $$0\to\Omega^1[-1]\to \Omega^{\bullet}_{alg} \to {\cal O} \to 0,$$ with the 'abuse of notation' of identifying a one-term complex with its corresponding term. (Note that this makes sense as a sequence of complexes - it wouldn't make sense to swap the outer [non-zero] terms.)

Taking (hyper-)cohomology gives the long exact sequence $$\cdots \to H^1(X, {\Omega}^1[1])\to {\mathbb H}^1(X, \Omega^{\bullet}_{alg}) \to H^1(X, {\cal O}) \to \\ H^2(X, \Omega^1[-1]) \to {\mathbb H}^2(X, \Omega^{\bullet}_{alg}) \to H^2(X, {\cal O}) \to \cdots.$$

But, on the one hand, $$H^k(X, \Omega^1[-1]) = H^{k-1}(X, \Omega^1)$$. On the other, in the case of $$X={\mathbb P}^1(\mathbb C)$$, $$H^1(X, {\cal O})$$ and $$H^2(X,{\cal O})$$ vanish. [Once again, using Hodge theory, we wouldn't need to rely on this]. Therefore one can identify $${\mathbb H}^2(X, \Omega^{\bullet}_{alg}) = H^1(X, \Omega^1) .$$ So, with $$X = {\mathbb P}^1({\mathbb C})$$, in terms of Cech cohomology, $$H^1(X, \Omega^1)$$ is spanned by (the class of) $$dz/z$$ on the open set $${\mathbb P^1}({\mathbb C}) \setminus \{0, \infty \}$$.

In the above, I was taking the hypercohomology of a complex of sheaves $$\cal C$$ to be (up to unique isomorphism) the homology of the complex of modules of global sections $$\Gamma(X, {\cal I})$$, where $${\cal I}$$ is a complex of injectives quasi-isomorphic to $${\cal C}$$.

This answer is obviously not very satisfactory. Still, hoping it was of some help... If someone feels like cleaning this up, or improving/rewriting it, rather than writing their own answer, I am happy to make this a community answer.

Comment/Edit: it's worth notating explicitly that G-H's $$\partial$$-Poincare statement $${\cal H}^k(\Omega^{\bullet}) =0$$ for $$k>0$$ does NOT hold in the Zariski topology. For instance $$dz / z$$ does not have an algebraic anti-derivative on any open set.

• Thank you for your detailed answer! I accepted the more algebraic one because that is more directly how I was thinking of this problem, but your answer really completes the picture for me. I wish I could accept both! Mar 28, 2019 at 9:58
• Also, quick question, does any of this extend in some obvious way to the other example I mentioned - the smooth hypersurface in $\mathbb{P_{\mathbb{C}}^3}$? This isn't a Riemann surface though. Mar 28, 2019 at 10:05
• I'm actually on the road for the next few days. I'll try to get back by the end of NEXT week. Mar 28, 2019 at 12:14
• Sure, thank you! Mar 28, 2019 at 14:36